Calculate the pH of an aqueous solution obtained by mixing 50ml of 0.2 M HCl with 50ml 0.1 M NaOH

Explanation

No of moles of HCl `= 0.2 xx 50 xx 10^(-3) = 10 xx 10^(-3)`
No of moles of NaOH `= 0.1 xx 50 xx 10^(-3) = 5 xx 10^(-3)`
No of moles of HCl after mixing `= 10 xx 10^(-3) - 5 xx 10^(-3)`
`= 5 xx10^(-3)`
after mixing total volume = 100mL
`:.` Concentration of HCl in moles per litre = `(5xx10^(-3) "mol")/(100xx10^(-3) L)`
`[H_3O^(+)] = 5xx10^(-2)M`
`pH = -log (5xx10^(-2))`
`= 2 - log 5`
`=2 - 0.6990`
`= 1.30`